Odds Ratio to Probability of Success

#1
We ran a logistic regression model with Passing the certification exam (0 or 1) as an outcome. We found that one of the strongest predictors is the student's program GPA, the highest the program GPA, the highest the odds of passing the certification exam.

Standardized GPA: p-value < .0001, B estimate = 1.7154, odds ratio = 5.559

I interpret this as, with every 0.33 unit (one standard deviation) increase in GPA, the odds of succeeding in the certification exam increased by 5.559 times.

However, clients want to understand this in terms of probability. I calculated probability by:

(5.559 - 1) x 100 = 455.9 percent

I'm having trouble explaining this percentage to our client. I thought probability of success is only supposed to range from 0 to 1. Any help would be appreciated!
 

hlsmith

Not a robit
#2
Yes, probabilities are contained within 0.0-1.0.


You are working with log odds not probabilities. You need to Expit your data into predicted probabilities. Give me a couple of moments and I will post a solution.
 
#4
Here is your probability:


log_odds = 1.7154
X = (exp(log_odds))/(1+(exp(log_odds)))
X = 0.848
Thanks so much! How do I explain the 0.848 in terms of my research question? With every standard deviation increase in GPA, the probability of passing the certification exam is now 84.8%????
 

hlsmith

Not a robit
#5
Close. As you know interpreting these problems can be difficult.


My attempt would be:


Students with a 1 unit (standard deviation) higher GPA have an 84% greater predicted probability of passing the exam.
 
#6
Close. As you know interpreting these problems can be difficult.


My attempt would be:


Students with a 1 unit (standard deviation) higher GPA have an 84% greater predicted probability of passing the exam.
Thanks so much! What happens if the log odds is negative? For instance, the models shows that being an economically disadvantaged is negatively associated with passing the certification exam.

economically disadvantaged (reference group = not economically disadvantaged): B = -1.0709, (EXP(B))/(1+(EXP(B))) = 25.5%. Can I say the chances decreased by 25.5% for students who are economically disadvantaged compared to those who are not?
 
#7
You will need both the intercept and the regression coefficient beta.

Your model is:

log(p/(1-p)) = intercept + beta*x

That will give:

p = exp(intercept + beta*x)/(1 + exp(intercept + beta*x))

Here are a few examples:

Code:
###
## This is an R program

(exp(0.5 + 1.7154*0))/(1+exp(0.5 + 1.7154*0)) #intercept= 0.5 , beta =1.7154, x = 0
# 0.6224593

(exp(0  + 1.7154*0))/(1+exp(0  + 1.7154*0)) #intercept= 0  , beta =1.7154, x = 0
#  0.5

(exp(0.5 + 1.7154*1))/(1+exp(0.5 + 1.7154*1)) #intercept= 0.5 , beta =1.7154, x = 1 
# 0.9016239

(exp(-0.5 + 1.7154*0))/(1+exp(-0.5 + 1.7154*0)) #intercept= -0.5 , beta =1.7154, x = 0 
# 0.3775407

(exp(-0.5 + 1.7154*1))/(1+exp(-0.5 + 1.7154*1)) #intercept= -0.5 , beta =1.7154, x = 1 
# 0.771253 

#what happens with a negative beta?
(exp(-0.5  -1.7154*1))/(1+exp(-0.5  -1.7154*1)) #intercept= -0.5 , beta = -1.7154, x = 1 
#  0.09837606

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